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Question
Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.
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Solution
Diameter of the Earth = 12800 km
So, radius of the Earth, R = 6400 km = 6.4 × 106 m
Time period of revolution of the Earth about its axis :
\[T = 24 \text{hr} = 24 \times 3600 \text{s}\]
\[\text{v} = \frac{2\pi r}{T}=\frac{2 \times 3 . 14 \times 64 \times {10}^6}{24 \times 3600}\]
\[\Rightarrow \text{v} = 465 . 185 \text{ m/s}\]
\[\text { Acceleration of the particle }: \]
\[a = \frac{v^2}{R} = \frac{\left( 465 . 185 \right)^2}{64 \times {10}^5} = 0 . 038 \text{ m/ s}^2\]
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